In the redox ionic equation ${\mathrm{MnO}}_4{}^{-}+{\mathrm{I}}^{-}+{\mathrm{H}}_2\mathrm{O}⟶2{\mathrm{MnO}}_2+{\mathrm{BrO}}_3{}^{-}+2{\mathrm{OH}}^{-}$

The equivalent weight of potassium permanganate of molar mass ' ${\mathrm{M}}^{\text{'}}\mathrm{g}{\mathrm{mol}}^{-1}$ is

The oxidation state of Mn in ${{\mathrm{MnO}}_4}^{-}$is $+7$

The oxidation state of Mn in ${\text{MnO}}_2$ is $+4$

The reaction is an reduction reaction, the net change in oxidation state is $3$ due to gain of $3$ electrons

$\mathrm{Equivalent} \mathrm{weight} = \mathrm{Mass}/\mathrm{no}.\mathrm{of} \mathrm{electrons} \mathrm{participate} \mathrm{in} \mathrm{reaction}$

Therefore equivalent weight of KMnO₄ $=\mathrm{M}/3$.